# INEQUALITY'S BLOG

## December 22, 2010

### A generalization of Chebyshev inequality

Filed under: Chebyshev's inequality — KKKVVV @ 6:44 pm

### Problem 075: Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 8:11 am

Let $a, b, c$ are positive real numbers. Prove that:

$\displaystyle \frac{a^2}{\sqrt{b+c}}+\frac{b^2}{\sqrt{c+a}}+\frac{c^2}{\sqrt{a+b}}\geq \frac{(a^2+b^2+c^2)\sqrt{a+b+c}}{\sqrt{2(ab+bc+ca})}$

## December 16, 2010

### Inequality 06: VAN KHEA

Filed under: Van Khea 06 — KKKVVV @ 5:22 pm

### Problem 74: van khea

Filed under: Uncategorized — KKKVVV @ 10:35 am

Let $a, b, c>0$. Prove that:  $\displaystyle \frac{ab}{\sqrt{c+a}}+\frac{bc}{\sqrt{a+b}}+\frac{ca}{\sqrt{b+c}}\leq \frac{\sqrt{2}}{\sqrt{3}}(a+b+c)^{\frac{3}{2}}-\frac{3}{\sqrt{2}}\sqrt{abc}$

### Problem 73: van khea

Filed under: Uncategorized — KKKVVV @ 10:31 am

Let $a, b, c>0$. Prove that $\displaystyle \biggl(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\biggl)\biggl(a^2+b^2+c^2-\frac{3abc}{a+b+c}\biggl)\geq ab+bc+ca$

## December 15, 2010

### Problem 05: VAN KHEA

Filed under: the new math,Van Khea 05 — KKKVVV @ 5:51 pm

### THE INEQUALITY 03: VAN KHEA

Filed under: the new math,Van Khea 03 — KKKVVV @ 2:41 pm

### A GENERALIZATION OF VANKHEA’S INEQUALITY

Filed under: the new math,Van Khea 04 — KKKVVV @ 10:03 am

### A GENERALIZATION OF CAUCHY – SCHWARZ’S INEQUALITY

Filed under: Cauchy-Schwarz's inequality — KKKVVV @ 7:19 am

## December 13, 2010

### Inequality

Filed under: the new math — KKKVVV @ 3:12 pm

## December 11, 2010

### YOUNG’S INEQUALITY

Filed under: Young's inequality — KKKVVV @ 4:04 pm

### A GENERALIZATION OF AM – GM INEQUALITY

Filed under: AM-GM inequality — KKKVVV @ 4:02 pm

### A GENERALIZATION OF HOLDER’S INEQUALITY

Filed under: Holder's inequality — KKKVVV @ 3:59 pm

# A GENERALIZATION OF HOLDER’S INEQUALITY

## BY VAN KHEA

### CAMBODIAN BUILDING ENGINEER

December 12 , 2010

Email: van_khea@yahoo.com

Let $\displaystyle p, q>0 ; \frac{1}{p}+\frac{1}{q}\leq 1$ and let $\displaystyle a_i\geq 0 ; \sum_{i=1}^{n}\alpha_i=1$ then

$\displaystyle \sum_{i=1}^{n}\alpha_i |a_ib_i|\leq \biggl(\sum_{i=1}^{n}\alpha_i |a_i|^p\biggl)^{\frac{1}{p}}\biggl(\sum_{i=1}^{n}\alpha_i |b_i|^q\biggl)^{\frac{1}{q}} ; \forall{(a_1, a_2, ..., a_n) ; (b_1, b_2, ..., b_n)\in \mathbb{R}^n}$ or $\mathbb{C}^n$

Proof

Theorem: (VAN KHEA): Let $a_i\geq 0$ and $x_i, y_i, z_i>0 ; (i=1, 2, ..., n)$ and let $k\geq 1 ; \alpha , \beta , \gamma\geq 0$ such that $k-\alpha-\beta-\gamma=1$ then

$\displaystyle \frac{a_1^k}{x_1^{\alpha}y_1^{\beta}z_1^{\gamma}}+...+\frac{a_n^k}{x_n^{\alpha}y_n^{\beta}z_n^{\gamma}}\geq \frac{(a_1+...+a_n)^k}{(x_1+...+x_n)^{\alpha}(y_1+...+y_n)^{\beta}(z_1+...+z_n)^{\gamma}}$

From above theorem  we have

$\displaystyle \sum_{i=1}^{n}\alpha_i |a_ib_i|=\sum_{i=1}^{n}\frac{(\alpha_i |a_ib_i|)^2}{\alpha_i^{1-\frac{1}{p}-\frac{1}{q}}(\alpha_i|a_i|^p)^{\frac{1}{p}}(\alpha_i|b_i|^q)^{\frac{1}{q}}}$$\displaystyle \geq \frac{(\sum_{i=1}^{n}\alpha_i |a_ib_i|)^2}{(\sum_{i=1}^{n}\alpha_i)^{1-\frac{1}{p}-\frac{1}{q}}(\sum_{i=1}^{n}\alpha_i |a_i|^p)^{\frac{1}{p}}(\sum_{i=1}^{n}\alpha_i |b_i|^q)^{\frac{1}{q}}}$

$\displaystyle \Leftrightarrow 1\geq \frac{\sum_{i=1}^{n}\alpha_i |a_ib_i|}{(\sum_{i=1}^{n}\alpha_i |a_i|^p)^{\frac{1}{p}}(\sum_{i=1}^{n}\alpha_i |b_i|^q)^{\frac{1}{q}}}$

$\Rightarrow$$\displaystyle \sum_{i=1}^{n}\alpha_i |a_ib_i|\leq \biggl(\sum_{i=1}^{n}\alpha_i |a_i|^p\biggl)^{\frac{1}{p}}\biggl(\sum_{i=1}^{n}\alpha_i |b_i|^q\biggl)^{\frac{1}{q}} ; \forall{(a_1, a_2, ..., a_n) ; (b_1, b_2, ..., b_n)\in \mathbb{R}^n}$ or $\mathbb{C}^n$

## Special cases

• If $\displaystyle \frac{1}{p}+\frac{1}{q}\leq 1$ and $\displaystyle \alpha_1=\alpha_2=...=\alpha_n=\frac{1}{n}$ then:

$\displaystyle \frac{1}{n}\sum_{i=1}^{n}|a_ib_i|\leq \biggl(\frac{1}{n}\sum_{i=1}^{n}|a_i|^p\biggl)^{\frac{1}{p}}\biggl(\frac{1}{n}\sum_{i=1}^{n}|b_i|^q\biggl)^{\frac{1}{q}}$

• If $\displaystyle \frac{1}{p}+\frac{1}{q}=1$ and $\displaystyle \alpha_1=\alpha_2=...=\alpha_n$ then:

$\displaystyle \sum_{i=1}^{n}|a_ib_i|\leq \biggl(\sum_{i=1}^{n}|a_i|^p\biggl)^{\frac{1}{p}}\biggl(\sum_{i=1}^{n}|b_i|^q\biggl)^{\frac{1}{q}}$